PAST QUESTIONS 2024

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Try the questions first, using not more than 15 minutes for each question, and watch the accompanying videos to see how the questions are solved.

Question 1

1. A fair die and a fair coin are thrown together once.

   \(\hspace{0.7cm} i)\) Write down the set of all possible outcomes.

   \(\hspace{0.5cm} ii)\) Find the probability of obtaining a prime number and a tail.

Show Solution

Question 1.a.i

Fair die sample space

\(\Rightarrow \{1, 2, 3, 4, 5, 6\} \)

Fair coin sample space

\(\Rightarrow \{\text{Head (H), Tail (T)}\} \)

Sample space for all possible outcomes:





Question 1.a.ii

Total possible outcomes = 12

Successful outcomes (S)

\(\Rightarrow\) Prime number and a tail

\(\Rightarrow\) {T2, T3, T5}

\(n(\text{successful outcomes}) = 3\)

\(Probability = \dfrac{n(\text{successful outcomes})}{\text{total outcomes}}\)

\(\Rightarrow Prob(S) = \dfrac{3}{12}\)

\(\Rightarrow Prob(S) = \dfrac{1}{4}\)

\(\therefore\) the probability of obtaining a prime number and a tail is \(\frac{1}{4}\)





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2. The map of a field is drawn to a scale of \(1:100\). If the width and area of the field on the map are 8 cm and 88 cm\(^2\) respectively, find in m\(^2\), the area of the actual field.

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Question 1.b.

Scale \(\Rightarrow 1:100\)

width \(\Rightarrow 8\) cm

\(\Rightarrow\) Actual width \(= 8 \ cm \times 100\)

\(\Rightarrow\) Actual width \(= 800 \ cm\)

\(\Rightarrow\) Actual width \(= 8 \ m\)

Area \(\Rightarrow 88\) cm\(^2\)

Area = length \(\times\) width

\(\Rightarrow \text{length} = \dfrac{\text{Area}}{\text{width}}\)

\(\Rightarrow \text{length} = \dfrac{88 \ cm^2}{8 \ cm}\)

\(\Rightarrow \text{length} = 11 \ cm\)

\(\Rightarrow\) Actual length \(= 11 \ cm \times 100\)

\(\Rightarrow\) Actual length \(= 1100 \ cm\)

\(\Rightarrow\) Actual length \(= 11 \ m\)

\(\Rightarrow\) Actual Area \(=\) Actual length \(\times\) Actual width

\(\Rightarrow\) Actual Area \( 11 \ m \times 8 \ m\)

\(\Rightarrow\) Actual Area \(= 88 \ m^2\)

\(\therefore\) the actual area of the field is 88 m\(^2\)





3. Copy and complete the \(3 \times 3\) magic square such that the sum of the numbers in each row, column and diagonal is equal to 21.

   
   
Show Solution

Question 1.c.

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Question 2

1. Given the vectors \(\mathbf{p} = \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix}\), \(\mathbf{q} = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix}\) and \(\mathbf{p} = \mathbf{q}\), find the values of \(m\) and \(n\).


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Solution

\(\mathbf{p} = \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix}\)

\(\mathbf{q} = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix}\)

\(\mathbf{p} = \mathbf{q}\)

Finding \(m\):

\(\Rightarrow \begin{pmatrix} m + 3 \\ 2 - n \end{pmatrix} = \begin{pmatrix} 3m - 1 \\ n - 8 \end{pmatrix}\)

\(\Rightarrow m + 3 = 3m - 1\)

\(\Rightarrow 3 + 1 = 3m - m\)

\(\Rightarrow 4 = 2m\)

\(\Rightarrow \dfrac{2m}{2} = \dfrac{4}{2}\)

\(\Rightarrow m = 2\)




Finding \(n\):

\(\Rightarrow 2 - n = n - 8\)

\(\Rightarrow 2 + 8 = n + n\)

\(\Rightarrow 10 = 2n\)

\(\Rightarrow \dfrac{2n}{2} = \dfrac{10}{2}\)

\(\Rightarrow n = 5\)

\(\therefore\) m is 2 and n is 5





2. A man shared an amount of money between his children Baaba and William in the ratio \(6:5\). Baaba received Gh₵1,200.00.

   \((i)\) Find the total amount shared.

   \((ii)\) William invested his share in an account at the rate of \(20\%\) simple interest per annum for 2 years. Find the total amount in his account at the end of the 2 years.


Show Solution

Question 2.b.i

Baaba's share \(=\) Gh₵ \(1,200.00\)

Baaba's ratio \(=\) 6

William's ratio \(=\) 5

Total ratio \(=\) \(6 + 5\)

\(\hspace{1.4cm}=\) \(11\)




If Baaba's ratio, 6 \(=\) Gh₵ \(1,200.00\)

then total ratio, 11 \(=\) \(\dfrac{11}{6} \times 1200\)

\(\Rightarrow\) \(11 \times 200\)

\(\Rightarrow\) Gh₵ \(2,200.00\)

\(\therefore\) the total amount shared was Gh₵ 2,200.00





Question 2.b.ii

Principal = William's share

William's share

\(\Rightarrow\) Amount shared \(-\) Baaba's share

\(\Rightarrow\) Gh₵ \(2,200\) \(-\) Gh₵ \(1,200\)

\(\Rightarrow\) Gh₵ \(1,000\)

\(\therefore\) Principal, P \(=\) Gh₵ \(1,000.00\)

Rate, R \(=\) \(20\%\)

Time, T \(=\) \(2\) years

Simple Interest, \(I = \dfrac{P \times R \times T}{100}\)

\(I = \dfrac{1000 \times 20 \times 2}{100}\)

\(I = 10 \times 40\)

\(I = 400\)

\(\therefore\) the interest earned was Gh₵ \(400.00\)




Total amount \(=\) Principal \(+\) Interest

Total amount \(= 1000 + 400\)

Total amount \(= 1,400.00\)

\(\therefore\) the total amount of Williams money was Gh₵ 1,400.00

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Question 3

1. Simplify: \(3\sqrt{50} + 2\sqrt{45} - \sqrt{2} + \sqrt{5}\).

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Solution

\(3\sqrt{50} + 2\sqrt{45} - \sqrt{2} + \sqrt{5}\)

\(3(5\sqrt{2}) + 2(3\sqrt{5}) - \sqrt{2} + \sqrt{5}\)

\(\Rightarrow 15\sqrt{2} + 6\sqrt{5} - \sqrt{2} + \sqrt{5}\)

Grouping like terms:

\(\Rightarrow 15\sqrt{2} - \sqrt{2} + 6\sqrt{5} + \sqrt{5}\)

\(\Rightarrow 14\sqrt{2} + 7\sqrt{5}\)





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Watch the video below for a walk through of the solution.











2. A wire of length 38 cm is bent into the shape of a rectangle whose length is 7 cm more than the width. Find the area of the rectangle.

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Question 3b

Let the width of the rectangle, W \(\Rightarrow x\).

Then the length of the rectangle, L \(\Rightarrow x + 7\).

The perimeter of the rectangle \(\Rightarrow 38\) cm.

\(\text{Perimeter} = 2(\text{L} + \text{W})\)

Substituting the values:

\(\Rightarrow 38 = 2(2x + 7)\)

\(\Rightarrow 38 = 4x + 14\)

\(\Rightarrow 38 - 14 = 4x\)

\(\Rightarrow 4x = 24\)

\(\Rightarrow \dfrac{4x}{4} = \dfrac{24}{4}\)

\(\Rightarrow x = 6\)

\(\therefore\) The width of the rectangle is \(6\) cm, and the length is \(6 + 7 = 13\) cm.

The area of the rectangle \(= \text{length} \times \text{width}\).

\(\Rightarrow \text{Area} = 13 \times 6\)

\(\Rightarrow \text{Area} = 78\) cm\(^2\)

\(\therefore\) The area of the rectangle is \(78\) cm\(^2\).




3. If \(15\%\) of the length of a rope is 720 metres, find half of the length of the rope.

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Question 3c

Half of rope \(= 50\%\)

If \(15\% = 720 \ \text{metres}\)

Then \(50\% = \dfrac{50}{15} \times 720 \ \text{metres}\)

\(\hspace{1.5cm}\Rightarrow \dfrac{10}{3} \times 720 \ \text{metres}\)

\(\hspace{1.5cm}\Rightarrow 10 \times 240 \ \text{metres}\)

\(\hspace{1.5cm}\Rightarrow 2,400 \ \text{metres}\)

\(\therefore\) Half of the length of the rope is 2,400 metres.

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Question 4

1. Using a ruler and a pair of compasses only, construct \(\triangle PQR\) such that angle \(PQR = 90^\circ\), \(|PQ| = 5.5\) cm and \(|QR| = 8\) cm.


2. Construct a perpendicular of \(\overline{PR}\) from \(Q\).


3. Locate \(M\), the intersection of the perpendicular and \(\overline{PR}\).


4. Measure:
   

   \((i)\hspace{0.65cm}\) \(|MR|\);

   \((ii)\hspace{0.5cm}\) \(|QM|\).



5. Calculate, correct to the nearest whole number, the area of triangle \(QMR\).

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Solution

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Question 5

1. In the diagram, \(\triangle PQR\) is an enlargement of \(\triangle PST\). \(|PS| = 4\) cm, \(|QS| = 2\) cm and \(|QR| = 10\) cm.

   \((i)\hspace{0.65cm}\) Find the length of \(\overline{ST}\).

   Show Solution

   Question 5.a.i

   \(|PS| = 4\) cm

   \(|QS| = 2\) cm

   \(|QR| = 10\) cm

   Scale factor, \(k = \dfrac{\text{image length}}{\text{object length}}\)

   \(\Rightarrow k = \dfrac{4 \ cm + 2 \ cm}{4 \ cm}\)

   \(\Rightarrow k = \dfrac{6 \ cm}{4 \ cm}\)

   \(\Rightarrow k = \dfrac{3}{2}\)

   \(\therefore\) the scale factor of the enlargement is \(\frac{3}{2}\)

   Scale factor, \(k = \dfrac{|\overline{QR}|}{|\overline{ST}|}\)

   \(\Rightarrow \dfrac{3}{2} = \dfrac{10 \ cm}{|\overline{ST}|}\)

   Cross multiply:

   \(\Rightarrow 3 \times |\overline{ST}| = 2 \times 10 \ cm\)

   \(\Rightarrow 3|\overline{ST}| = 20 \ cm\)

   \(\Rightarrow \dfrac{3|\overline{ST}|}{3} = \dfrac{20}{3} \ cm\)

   \(\Rightarrow |\overline{ST}| = \dfrac{20}{3} \ cm\)

   \(\Rightarrow |\overline{ST}| = 6.6667 \ cm\)

   \(\therefore\) the length of \(\overline{ST}\) is 6.6667 cm

   
   

   \((ii)\hspace{0.5cm}\) If \(|\overline{PQ}| = |\overline{PR}|\), find the area of \(\triangle PQR\).

   Show Solution

   Question 5.a.ii

   

   \(|\overline{PQ}| = |\overline{PR}|\)

   let \(M \Rightarrow\) the midpoint of \(\overline{QR}\)

   \(\Rightarrow |\overline{PM}|\) is perpendicular to \(|\overline{QR}|\)

   \(\Rightarrow |\overline{MR}| = \frac{1}{2}|\overline{QR}|\)

   \(\Rightarrow |\overline{MR}| = \frac{1}{2}\times 10 \ cm\)

   \(\Rightarrow |\overline{MR}| = 5 \ cm\)

   Using the Pythagoras theorem:

   \(|\overline{PR}|^2 = |\overline{MR}|^2 + |\overline{PM}|^2 \)

   \(\Rightarrow 6^2 = 5^2 + |\overline{PM}|^2\)

   \(\Rightarrow |\overline{PM}|^2 = 36 - 25 \)

   \(\Rightarrow |\overline{PM}|^2 = 11 \)

   \(\Rightarrow |\overline{PM}| = \sqrt{11} \ cm \)

   Area of triangle

   \(\Rightarrow \frac{1}{2} \times \text{length} \times \text{width} \)

   Area of \(\triangle PQR\)

   \(\Rightarrow \frac{1}{2} \times |\overline{QR}| \times |\overline{PR}|\)

   \(\Rightarrow \frac{1}{2} \times 10 \ cm \times \sqrt{11} \ cm\)

   \(\Rightarrow 5 \ cm \times \sqrt{11} \ cm\)

   \(\Rightarrow 5\sqrt{11} \ cm^2\)

   \(\therefore\) the area of \(\triangle PQR\) is \(5\sqrt{11} \ cm^2\)




2. The total area of a school compound is 900\(\frac{1}{2}\) m\(^2\). The school has Administration and Classroom block, Library, School Park, Roads and Walkways. The areas of the Administration and Classroom block, Library and School Park are 300\(\frac{1}{4}\) m\(^2\), 200\(\frac{1}{2}\) m\(^2\), and 120\(\frac{1}{8}\) m\(^2\) respectively. Find the area covered by Roads and Walkways altogether.

Show Solution

Question 5.b

Total area of school compound \( \Rightarrow 900\frac{1}{2}\) m\(^2\)

Administration and classroom block \( \Rightarrow 300\frac{1}{4}\) m\(^2\)

Area covered by Library \(\Rightarrow 200\frac{1}{2}\) m\(^2\)

Area covered by School Park \(\Rightarrow 120\frac{1}{8}\) m\(^2\)

Hence, area covered by Roads and Walkways:

\(\Rightarrow\) \(900\frac{1}{2} - \left( 300\frac{1}{4} + 200\frac{1}{2} + 120\frac{1}{8}\right)\)

\(\Rightarrow\) \(\frac{1801}{2} - \left(\frac{1201}{4} + \frac{401}{2} + \frac{961}{8}\right)\)

\(\Rightarrow\) \(\frac{1801}{2} - \left(\frac{2(1201) + 4(401) + 1(961)}{8}\right)\)

\(\Rightarrow\) \(\frac{1801}{2} - \left(\frac{2402 + 1604 + 961}{8}\right)\)

\(\Rightarrow\) \(\frac{1801}{2} - \frac{4967}{8}\)

\(\Rightarrow\) \(\frac{4(1801) - 1(4967)}{8}\)

\(\Rightarrow\) \(\frac{7204 - 4967}{8}\)

\(\Rightarrow\) \(\frac{2237}{8}\)

\(\Rightarrow\) \(279\frac{5}{8}\) m\(^2\)

\(\therefore\) the area covered by Roads and Walkways is \(279\frac{5}{8}\) m\(^2\)

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Question 6

1. Copy and complete the table for the relation \(F = \frac{9}{5}C + 32\),
   where \(F\) and \(C\) are degrees Fahrenheit and degrees Celsius respectively.
   

   


2. Using a scale of 2 cm to 10 units on the vertical axis \((^\circ F)\) and 2 cm to 5 units on the horizontal axis \((^\circ C)\), draw a linear graph for the relation.


3. Use the graph to find the temperature in degrees celcius when \(F = 55\) degrees.


4. Interpret the slope of the relation.

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Solution

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